Question: What is the extraneous solution to these equations? $\dfrac{x^2 - 3x}{x + 2} = \dfrac{-13x - 16}{x + 2}$
Multiply both sides by $x + 2$ $ \dfrac{x^2 - 3x}{x + 2} (x + 2) = \dfrac{-13x - 16}{x + 2} (x + 2)$ $ x^2 - 3x = -13x - 16$ Subtract $-13x - 16$ from both sides: $ x^2 - 3x - (-13x - 16) = -13x - 16 - (-13x - 16)$ $ x^2 - 3x + 13x + 16 = 0$ $ x^2 + 10x + 16 = 0$ Factor the expression: $ (x + 2)(x + 8) = 0$ Therefore $x = -2$ or $x = -8$ At $x = -2$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -2$, it is an extraneous solution.